Let $g(x)=e^{\tan(x)}$. Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\tan(x)e^{\tan(x)-1}$ (Choice B) B $e^{\tan(x)}\sec^2(x)$ (Choice C) C $e^{\tan(x)}$ (Choice D) D $e^{\sec^2(x)}$
Solution: $g(x)$ is a composition of two, more basic, functions: $\tan(x)$ and $e^x$. In other words, suppose $u(x)=\tan(x)$ and $v(x)=e^x$, then $g(x)=v\Bigl(u(x)\Bigr)$, or $(v\circ u)(x)$. Therefore, the derivative of $g$ can be found using the chain rule : $\begin{aligned} \dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&=\dfrac{dv}{du}\cdot\dfrac{du}{dx} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x) \end{aligned}$ Finding $v'\Bigl(u(x)\Bigr)$ $v(x)=e^x$, and therefore $v'(x)=e^x$. Now we plug $u(x)=\tan(x)$ into $v'$ : $\begin{aligned} v'\Bigl(u(x)\Bigr)&=v'\Bigl(\tan(x)\Bigr) \\\\ &={e^{\tan(x)}} \end{aligned}$ Finding $u'(x)$ $u(x)=\tan(x)$, and therefore $u'(x)={\sec^2(x)}$. Putting things together $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left(e^{\tan{x}}\right) \\\\ &=\dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&&\gray{\text{Let }u(x)=\tan(x)\text{, }v(x)=e^x} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x)&&\gray{\text{The chain rule}} \\\\ &={e^{\tan(x)}}\cdot{\sec^2(x)} \\\\ &=e^{\tan(x)}\sec^2(x) \end{aligned}$ In conclusion, $g'(x)=e^{\tan(x)}\sec^2(x) $.